# Spectrum of a ring of continuous functions

The set $A=\{f:[0,1] \rightarrow \mathbb{R} : f$ is continuous$\}$ is a ring with the standard function addition and multiplication. Which are the prime ideals in $A$?

The only thing I've managed to observe is that the units of this ring are the functions that never take the value $0$, so every element of a prime ideal $P$ takes the value $0$ at least once.

• did you figure out which are the ideals in that ring? – Jorge Fernández Hidalgo Feb 13 '15 at 5:23
• I suspects the ideals are of the form $\{f \in A: f(x)=0 \forall x \in S\}$, where $S$ is a subset of $[0,1]$ – Marco Flores Feb 13 '15 at 6:24
• The maximal ideals are well-known. But what about the prime ideals? good question. – Mister Benjamin Dover Feb 14 '15 at 17:02

## 1 Answer

fix $x_0$ ,then ideal $I_0=\{f\in A: f(x_0)=0\}$ is prime. For that if $fg\in I_0$ Then either $f$ or $g$ must be zero at $x_0$.

Edit1:i dont know if these are all ...

• I can see those are prime ideals.. but, why are they all the prime ideals? – Marco Flores Feb 13 '15 at 6:01
• You get that every element $f$ of a proper ideal $I$ satisfies $f(x)=0$ for some $x$, but that $x$ need not be the same for every $f \in I$, or does it? – Marco Flores Feb 13 '15 at 6:09
• Those are all maximal ideals. – user26857 Feb 13 '15 at 7:06
• And so prime ideals.but apparently there are examples of prime ideals of C(I) that arent maximal. – BigM Feb 13 '15 at 7:09